What is the average value of $e^{x}$ on the interval $[3,9]$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{e^{9}+e^{3}}{6}$ (Choice B) B $\dfrac{e^{9}+e^{3}}{2}$ (Choice C) C $\dfrac{e^{9}-e^{3}}{6}$ (Choice D) D $\dfrac{e^{9}-e^{3}}{2}$
Explanation: In general, this is the average value of function $f$ over the interval $[a,b]$ : $\dfrac{\int_a^b f(x)\,dx}{b-a}$ In our case, ${f(x)=e^{x}}$, ${a=3}$ and ${b=9}$ : $\begin{aligned} \dfrac{\int_{ a}^{ b} {f(x)}\,dx}{ b- a}&=\dfrac{\int_{{3}}^{ {9}} ({e^{x}})\,dx}{{9}-{3}} \\\\ &=\dfrac{\Big[e^{x}\Big]_{3}^{9}}{6} \\\\ &=\dfrac{e^{9}-e^{3}}{6} \end{aligned}$ In conclusion, the average value of $e^{x}$ on the interval $[3,9]$ is $\dfrac{e^{9}-e^{3}}{6}$.